Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 46 / 2

The vertical hydrostatic equilibrium equation balances the gradient of the total pressure against gravity. In a thin disk with Keplerian rotation, expanding the central gravitational field about the midplane gives vertical acceleration $-{\Omega }^{2}z$, so $p_z=-\rho {\Omega }^{2}z$. Disk self-gravity is neglected in this approximation.

The flux-gradient equation is local thermal balance: the upward radiative flux increases with height because differential rotation dissipates energy. For a Newtonian effective dynamic viscosity $\mu$, the volumetric heating rate is $\mu (r{\Omega }^{\prime }{)}^2$, which becomes $9\mu {\Omega }^2/4$ for Keplerian rotation. The symbol $\mu$ here is dynamic viscosity, not kinematic viscosity; the latter is $\nu =\mu /\rho$.

The radiative diffusion equation applies to an optically thick medium in local thermodynamic equilibrium. Its diffusion coefficient is $c/(3\kappa \rho )$, and the equilibrium radiation energy density is $u_{\mathrm{rad}}=4\sigma T^4/c$. HenceThe minus sign means energy flows down the temperature gradient. Here $\sigma$ is the Stefan-Boltzmann constant, $\kappa$ the opacity per unit mass, and $c$ the speed of light. Constant Thomson scattering opacity is the adopted transport approximation; local thermodynamic equilibrium is an additional assumption behind the radiation equation of state.

Finally the equation of state adds ideal gas gas pressure $p_g=\rho kT/({\mu }_m{m}_H)$ and isotropic radiation pressure $p_r=u_{\mathrm{rad}}/3=4\sigma T^4/(3c)$. The Boltzmann constant is $k$, and ${\mu }_m{m}_H$ is the mean mass per gas particle. Both contributions supply the total pressure entering vertical hydrostatic equilibrium.

In the limiting model with negligible gas pressure, $p=p_r=4\sigma T^4/(3c)$. Its gradient is $p_z=16\sigma T^3{T}_z/(3c)$. Combining radiative diffusion and hydrostatic equilibrium givesBecause $\kappa$ and $\Omega$ are constant with height at the fixed disk radius,Equating this to the viscous heating rate $9\mu {\Omega }^2/4$ forcesThis proves constant viscosity in radiation-supported vertical balance. It fixes the local effective dynamic viscosity uniquely, independent of height, surface density and radius within this ideal limiting model. It does not fix a height-independent kinematic viscosity: $\nu (z)=4c/(9\kappa \rho (z))$ still depends on the local mass density.

Set $\mathcal{R}=k/({\mu }_m{m}_H)$, so $p_g=\mathcal{R}\rho T$ and the prescribed dynamic viscosity is $\mu =\alpha \mathcal{R}\rho T/\Omega$. Keplerian shear and radiative diffusion then giveDifferentiating the second expression, equating it to the first and dividing by $\rho$ givesThis reduction uses the gas-pressure viscosity closure, not an assumption that gas pressure dominates the hydrostatic support.

For $z\ge 0$, introduce the vertical mass coordinate of a diskLet $\Sigma ={\int }_{-z_s}^{z_s}\rho dz$ be the full surface density of a reflection-symmetric disk. The upper surface is at $m_s=\Sigma /2$. With $\zeta =2m/\Sigma$, the temperature equation becomesWrite $T=T_{\ast }t(\zeta )$ and chooseThe equation reduces exactly to $(t^4{)}^{\prime \prime }+t=0$. Midplane symmetry gives $t^{\prime }(0)=0$, and the zero surface temperature gives $t(1)=0$. The stipulated unique positive nontrivial dimensionless solution therefore applies at every radius and surface density, with all physical dependence contained in $T_{\ast }$ and the mass-coordinate scaling. A zero temperature at the ideal surface does not mean zero escaping flux: $F=-(4\sigma /(3\kappa )){\partial }_m{T}^4$ can have a finite nonzero surface value.

The required viscosity is the density-weighted mean kinematic viscosity, not an unweighted vertical mean. Since $\rho \nu =\mu$,Substituting the dimensionless solution gives, with $I_t={\int }_0^{1}t(\zeta )d\zeta$,For central mass $M_{\ast }$, $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$, so the requested explicit radius and surface-density dependence isThe universal integral need not be evaluated. No separate solution for $\rho (z)$ is needed to find this mean: the mass-coordinate transformation removes it from both the temperature equation and the viscosity integral. The result is the gas-pressure viscosity closure with mixed pressure support.